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Find voltage between two nodes

by builder_user
Tags: nodes, voltage
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builder_user
#1
Mar24-11, 03:38 PM
P: 196
1. The problem statement, all variables and given/known data
Find voltage between node 1 and node 2


2. Relevant equations
What is the algorithm?


3. The attempt at a solution
R1=4 Ohms
R2=5 Ohms
R3=8 Ohms
R4=3 Ohms
R5=6 Ohms
R6=7 Ohms
Eэ=20 V
E3=17 V
E5=15 V
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cupid.callin
#2
Mar24-11, 03:47 PM
P: 1,135
use balanced wheat stone bridge thing
builder_user
#3
Mar24-11, 03:55 PM
P: 196
Do I need second Kirhgoff rules' equations for every loop?

cupid.callin
#4
Mar24-11, 04:01 PM
P: 1,135
Find voltage between two nodes

try "balanced wheat stone bridge"

you wont need Kirhgoff rules

here:http://en.wikipedia.org/wiki/Wheatstone_bridge
tiny-tim
#5
Mar24-11, 04:04 PM
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P: 26,148
hi builder_user!
Quote Quote by builder_user View Post
Do I need second Kirhgoff rules' equations for every loop?
(Kirchhoff!)

'fraid so!
builder_user
#6
Mar24-11, 04:20 PM
P: 196
What's the next step after Kirchgoff's rules?What do I need to find?I know current in every branch.
gneill
#7
Mar24-11, 04:29 PM
Mentor
P: 11,858
You want the voltage between nodes 1 and 2. Use the current you found for the first loop (leftmost) using Kirchhoff to determine the voltage across R1. Then do the obvious voltage sum.
builder_user
#8
Mar24-11, 04:35 PM
P: 196
i1=3.7A

U=Eэ-i1*R1=5.2V?

And the final result -- 20(Eэ)+5.2(U)=25.2V?
gneill
#9
Mar24-11, 04:49 PM
Mentor
P: 11,858
That's not the value I get for i1. Better check your derivation.

EDIT: My error. Redoing my sums I see that i1 is indeed about 3.7A.
builder_user
#10
Mar24-11, 05:14 PM
P: 196
Quote Quote by gneill View Post
That's not the value I get for i1. Better check your derivation.
Ok.But what's the next?
gneill
#11
Mar24-11, 05:17 PM
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P: 11,858
Quote Quote by builder_user View Post
Ok.But what's the next?
The next what? Wasn't it the voltage between nodes 1 and 2 that you were looking for?
builder_user
#12
Mar24-11, 05:23 PM
P: 196
but i's only I1.
Or R1*I1 - is the result?
gneill
#13
Mar24-11, 05:24 PM
Mentor
P: 11,858
Very sorry. I see that your figure of 3.7A for i1 and about 5.2V for the voltage between nodes 1 and 2 look okay.
builder_user
#14
Mar24-11, 05:36 PM
P: 196
Quote Quote by gneill View Post
Very sorry. I see that your figure of 3.7A for i1 and about 5.2V for the voltage between nodes 1 and 2 look okay.
A-a-a-a-a!OMG!I'm already started to do all my work from the beginig...It's about 15 pages A4!
gneill
#15
Mar24-11, 05:46 PM
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P: 11,858
Again, sorry about that.

15 pages of A4? That seems like a lot for just three loops. What method are you using to solve the equations?
builder_user
#16
Mar24-11, 05:50 PM
P: 196
Quote Quote by gneill View Post
Again, sorry about that.

15 pages of A4? That seems like a lot for just three loops. What method are you using to solve the equations?
It's a big work like coursework.I have 8 exercises need to do with this scheme(different trasformations like triangle-star and different methods of finding currents like method of equivalent generator(I don't know how does it called in English.translate from russian),node voltage method and etc.).And another 7 schemes.One of them is my previous topic's scheme.

I got this result(3.7) by different methods.So if it's wrong - all my work is wrong
gneill
#17
Mar24-11, 06:04 PM
Mentor
P: 11,858
I see, so this is just one exercise for the given schematic.

It seems to me that a KVL loop approach is the most straight forward for this particular question, since you really only need to solve for the current in the first loop. Have you learned the method to directly write (by inspection) the matrix form of the equations? That allows you solve for the currents using a matrix method (such as Cramer's Rule), which takes only a few lines.
builder_user
#18
Mar24-11, 06:12 PM
P: 196
if it's cramer's method...
I know this method but I only use it in programming.I solve all equtations in MathCAd.And then I show result without numbers and then the final result.The result without numbers is very big sometimes.


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